JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given, the base of a prism is equilateral triangle of side 10 cm as shown in figure.
At minimum deviation, Angle of incident = Angle of prism
i.e. i = A = 60° Angle of refraction, r=

A

2

=60°2=30°
Let µ be the refractive index.
Then, by Snell’s law of refraction,
  µ=

sin‌i

sin‌r

=

sin‌60°

sin‌30°

=

√3 2

1 2

=√3
Now, in ∆ABP, AP = AB cos 30°
=

10√3

2

=5√3‌cm
=5√3×10−2m
i.e. optical distance travelled by light along AP,
d=µ×AP=√3×5√3×10−2=15×10−2m
Now, the time taken by light (c=3×108m∕s) to travel from P to A,
t=

d

C

=

15×10−2

3×108

s=5×10−10s
Thus, correct answer is 5.